EBK THERMODYNAMICS: AN ENGINEERING APPR
EBK THERMODYNAMICS: AN ENGINEERING APPR
8th Edition
ISBN: 9780100257054
Author: CENGEL
Publisher: YUZU
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Chapter 13.3, Problem 42P

a)

To determine

The volumetric flow rate and mass flow rate of the mixture treating it as an ideal gas mixture.

a)

Expert Solution
Check Mark

Answer to Problem 42P

The volume flow rate is 0.001005m3/s.

The mass flow rate is 0.09535kg/s.

Explanation of Solution

Write the expression to obtain the mass of O2 (mO2).

mO2=NO2MO2 (I)

Here, number of moles of O2 is NO2 and molar mass of O2 is MO2.

Write the expression to obtain the mass of N2 (mN2).

mN2=NN2MN2 (II)

Here, number of moles of N2 is NN2 and molar mass of N2 is MN2.

Write the expression to obtain the mass of CO2 (mCO2).

mCO2=NCO2MCO2 (III)

Here, number of moles of CO2 is NCO2 and molar mass of CO2 is MCO2.

Write the expression to obtain the mass of CH4 (mCH4).

mCH4=NCH4MCH4 (IV)

Here, number of moles of CH4 is NCH4 and molar mass of CH4 is MCH4.

Write the expression to obtain the total mass of each component (mm).

mm=mO2+mN2+mCO2+mCH4 (V)

Write the expression to obtain the total number of moles (Nm).

Nm=NO2+NN2+NCO2+NCH4 (VI)

Write the expression to obtain the molar mass of the mixture (Mm).

Mm=mmNm (VII)

Write the expression to obtain the gas constant of the mixture (R).

R=RuMm (VIII)

Here, universal gas constant is Ru.

Write the expression to obtain the specific volume of the mixture (v).

v=RTP (IX)

Here, temperature and pressure is TandP.

Write the expression to obtain the volume flow rate (V˙).

V˙=AV=πD24V (X)

Here, velocity is V, area is A, and diameter is D.

Write the expression to obtain the mass flow rate (m˙).

m˙=V˙v (XI)

Conclusion:

Refer Table A-1, “Molar mass, gas constant, and critical point properties”, obtain the molar masses of O2, N2, CO2, CH4 as 32.0kg/kmol, 28.0kg/kmol, 44.0kg/kmol, and 16.0kg/kmol respectively.

Substitute 30 kmol for NO2 and 32.0kg/kmol for MO2 in Equation (I).

mO2=(30kmol)(32.0kg/kmol)=960kg

Substitute 40 kmol for NN2 and 28.0kg/kmol for MN2 in Equation (II).

mN2=(40kmol)(28.0kg/kmol)=1,120kg

Substitute 10 kmol for NCO2 and 44.0kg/kmol for MCO2 in Equation (III).

mCO2=(10kmol)(44.0kg/kmol)=440kg

Substitute 20 kmol for NCH4 and 16.0kg/kmol for MCH4 in Equation (IV).

mCH4=(20kmol)(16.0kg/kmol)=320kg

Substitute 960 kg for mO2, 1120 kg for mN2, 440 kg for mCO2, and 320 kg for mCH4 in Equation (V).

mm=960kg+1120kg+440kg+320kg=2840kg

Substitute 30kmol for NO2, 40kmol for NN2, 10kmol for NCO2 and 20kmol for NCH4 in Equation (VI).

Nm=30kmol+40kmol+10kmol+20kmol=100kmol

Substitute 2,840 kg for mm and 100 kmol for Nm in Equation (VII).

Mm=2,840kg100kmol=28.40kg/kmol

Substitute 8.314kJ/kmolK for Ru and 28.40kg/kmol for Mm in Equation (VIII).

Rm=8.314kJ/kmolK28.40kg/kmol=0.2927kJ/kgK

Substitute 0.2927kJ/kgK for R, 15°C for T, and 8 MPa for P in Equation (IX).

v=(0.2927kJ/kgK)(15°C+273)K8MPa=(0.2927kJ/kgK)288K8MPa×1,000(kPa1MPa)=0.01054m3/kg

Substitute 0.016m2 for D, and 5m/s for V in Equation (X).

V˙=π(0.016m2)24(5m/s)=0.001005m3/s

Thus, the volume flow rate is 0.001005m3/s.

Substitute 0.001005m3/s for V˙ and 0.01054m3/kg for v in Equation (XI).

m˙=0.001005m3/s0.01054m3/kg=0.09535kg/s

Thus, the mass flow rate is 0.09535kg/s.

b)

To determine

The volumetric flow rate and mass flow rate using a compressibility factor based on Amagad’s law of additive volume.

b)

Expert Solution
Check Mark

Answer to Problem 42P

The volume flow rate is 0.001005m3/s.

The mass flow rate is 0.1095kg/s.

Explanation of Solution

Write the expression to obtain the reduced temperature of O2(TR,O2).

TR,O2=TmTcr,O2 (XII)

Here, critical temperature of O2 is Tcr,O2.

Write the expression to obtain the reduced pressure of O2(PR,O2).

PR,O2=PmPcr,O2 (XIII)

Here, critical temperature of O2 is Pcr,O2.

Write the expression to obtain the reduced temperature of N2(TR,N2).

TR,N2=TmTcr,N2 (XIV)

Here, critical temperature of N2 is Tcr,N2.

Write the expression to obtain the reduced pressure of N2(PR,N2).

PR,N2=PmPcr,N2 (XV)

Here, critical temperature of N2 is Pcr,N2.

Write the expression to obtain the reduced temperature of CO2(TR,CO2).

TR,CO2=TmTcr,CO2 (XVI)

Here, critical temperature of CO2 is Tcr,CO2.

Write the expression to obtain the reduced pressure of CO2(PR,CO2).

PR,CO2=PmPcr,CO2 (XVII)

Here, critical temperature of CO2 is Pcr,CO2.

Write the expression to obtain the reduced temperature of CH4(TR,CH4).

TR,CH4=TmTcr,CH4 (XVIII)

Here, critical temperature of CH4 is Tcr,CH4.

Write the expression to obtain the reduced pressure of CH4(PR,CH4).

PR,CH4=PmPcr,CH4 (XIX)

Here, critical temperature of CH4 is Pcr,CH4.

Write the expression to obtain the compressibility factor of a mixture (Zm).

Zm=yiZi=yO2ZO2+yN2ZN2+yCO2ZCO2+yCH4ZCH4 (XX)

Here, compressibility factor of O2 is ZO2, compressibility factor of N2 is ZN2, compressibility factor of CO2 is ZCO2, and compressibility factor of CH4 is ZCH4.

Write the expression to obtain the specific volume of the mixture (v).

v=ZmRTP (XXI)

Write the expression to obtain the volume flow rate (V˙).

V˙=AV=πD24V (XXII)

Here, velocity is V, area is A, and diameter is D.

Write the expression to obtain the mass flow rate (m˙).

m˙=V˙v (XXIII)

Conclusion:

Substitute 288 K for Tm and 154.8 K for Tcr,O2 in Equation (XII).

TR,O2=288K154.8K=1.860

Substitute 8 MPa for Pm and 5.08 MPa for Pcr,O2 in Equation (XIII).

PR,O2=8MPa5.08MPa=1.575

Refer Table A-15, “Nelson-Olbert generalized compressibility chart”, obtain compressibility factor, Z for O2 as 0.95.

Substitute 288 K for Tm and 126.2 K for Tcr,N2 in Equation (XIV).

TR,N2=288K126.2K=2.282

Substitute 8 MPa for Pm and 3.39 MPa for Pcr,N2 in Equation (XV).

PR,N2=8MPa3.39MPa=2.360

Refer Table A-15, “Nelson-Olbert generalized compressibility chart”, obtain compressibility factor, Z for N2 as 0.99.

Substitute 288 K for Tm and 304.2 K for Tcr,CO2 in Equation (XVI).

TR,CO2=288K304.2K=0.947

Substitute 8 MPa for Pm and 7.39 MPa for Pcr,CO2 in Equation (XVII).

PR,CO2=8MPa7.39MPa=1.083

Refer Table A-15, “Nelson-Olbert generalized compressibility chart”, obtain compressibility factor, Z for CO2 as 0.199.

Substitute 288 K for Tm and 191.1 K for Tcr,CH4 in Equation (XVIII).

TR,CH4=288K191.1K=1.507

Substitute 8 MPa for Pm and 4.64 MPa for Pcr,CH4 in Equation (XIX).

PR,CH4=8MPa4.64MPa=1.724

Refer Table A-15, “Nelson-Olbert generalized compressibility chart”, obtain compressibility factor, Z for CH4 as 0.85.

Substitute 0.30 for yO2, 0.95 for ZO2, 0.40 for yN2, 0.99 for ZN2, 0.10 for yCO2, 0.199 for ZCO2, 0.20 for yCH4, and 0.85 for ZCH4 in Equation (XX).

Zm=0.30(0.95)+0.40(0.99)+0.10(0.199)+0.20(0.85)=0.8709

Substitute 0.8709 for Zm, 0.2927kJ/kgK for R, 15°C for T, and 8 MPa for P in Equation (XXI).

v=0.8709(0.2927kJ/kgK)(15°C+273)K8MPa(103kPa1MPa)=0.009178m3/kg

Substitute 0.016m2 for D, and 5m/s for V in Equation (XXII).

V˙=π(0.016m2)24(5m/s)=0.001005m3/s

Thus, the volume flow rate is 0.001005m3/s.

Substitute 0.001005m3/s for V˙ and 0.009178m3/kg for v in Equation (XXIII).

m˙=0.001005m3/s0.009178m3/kg=0.1095kg/s

Thus, the mass flow rate is 0.1095kg/s.

c)

To determine

The volumetric flow rate and mass flow rate using Key’s pseudocritical pressure and temperature.

c)

Expert Solution
Check Mark

Answer to Problem 42P

The volume flow rate is 0.001005m3/s.

The mass flow rate is 0.1037kg/s.

Explanation of Solution

Write the expression to obtain the pseudo-critical temperature of the mixture (Tcr,m).

Tcr,m=yiTcr,i=yO2Tcr,O2+yN2Tcr,N2+yCO2Tcr,CO2+yCH4Tcr,CH4 (XXIV)

Here, critical temperature of O2,N2,CO2,andCH4 are Tcr,O2,Tcr,N2,Tcr,CO2,andTcr,CH4.

Write the expression to obtain the pseudo-critical pressure of the mixture (Pcr,m).

Pcr,m=yiPcr,i=yO2Pcr,O2+yN2Pcr,N2+yCO2Pcr,CO2+yCH4Pcr,CH4 (XXV)

Here, critical pressure of O2,N2,CO2,andCH4 are Pcr,O2,Pcr,N2,Pcr,CO2,andPcr,CH4.

Write the expression to obtain the reduced temperature (TR).

TR=TmTcr,m (XXVI)

Write the expression to obtain the reduced pressure (PR).

PR=PmPcr,m (XXVII)

Write the expression to obtain the specific volume of the mixture (v).

v=ZmRTP (XXVIII)

Write the expression to obtain the volume flow rate (V˙).

V˙=AV=πD24V (XXIX)

Write the expression to obtain the mass flow rate (m˙).

m˙=V˙v (XXX)

Conclusion:

Substitute 0.30 for yO2, 154.8 K for Tcr,O2, 0.40 for yN2, 126.2 K for Tcr,N2, 0.10 for yCO2, 304.2 K for Tcr,CO2, 0.20 for yCH4, and 191.1 K for Tcr,CH4 in Equation (XXIV).

Tcr,m=0.30(154.8K)+0.40(126.2K)+0.10(304.2K)+0.20(191.1K)=165.6K

Substitute 0.30 for yO2, 5.08 MPa for Pcr,O2, 0.40 for yN2, 3.39 MPa for Pcr,N2, 0.10 for yCO2, 7.39 MPa for Pcr,CO2, 0.20 for yCH4, and 4.64 MPa for Pcr,CH4 in Equation (XXV).

Pcr,m=0.30(5.08MPa)+0.40(3.39MPa)+0.10(7.39MPa)+0.20(4.64MPa)=4.547MPa

Substitute 288 K for Tm and 165.6 K for Tcr,m in Equation (XXVI).

TR=288K165.6K=1.739

Substitute 8 MPa for Pm and 4.547 MPa for Pcr,m in Equation (XXVII).

PR=8MPa4.547MPa=1.759

Refer Table A-15, “Nelson-Olbert generalized compressibility chart”, obtain compressibility factor, Zm as 0.92.

Substitute 0.92 for Zm, 0.2927kJ/kgK for R, 15°C for T, and 8 MPa for P in Equation (XXVIII).

v=0.92(0.2927kJ/kgK)(15°C+273)K8MPa(103kPa1MPa)=0.009694m3/kg

Substitute 0.016m2 for D, and 5m/s for V in Equation (XXIX).

V˙=π(0.016m2)24(5m/s)=0.001005m3/s

Thus, the volume flow rate is 0.001005m3/s.

Substitute 0.001005m3/s for V˙ and 0.009694m3/kg for v in Equation (XXX).

m˙=0.001005m3/s0.009694m3/kg=0.1037kg/s

Thus, the mass flow rate is 0.1037kg/s.

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Chapter 13 Solutions

EBK THERMODYNAMICS: AN ENGINEERING APPR

Ch. 13.3 - Prob. 11PCh. 13.3 - Prob. 12PCh. 13.3 - Prob. 13PCh. 13.3 - Is a mixture of ideal gases also an ideal gas?...Ch. 13.3 - Express Daltons law of additive pressures. Does...Ch. 13.3 - Express Amagats law of additive volumes. Does this...Ch. 13.3 - How is the P-v-T behavior of a component in an...Ch. 13.3 - Prob. 18PCh. 13.3 - Prob. 19PCh. 13.3 - Prob. 20PCh. 13.3 - Prob. 21PCh. 13.3 - Consider a rigid tank that contains a mixture of...Ch. 13.3 - Is this statement correct? The volume of an...Ch. 13.3 - Is this statement correct? The temperature of an...Ch. 13.3 - Is this statement correct? The pressure of an...Ch. 13.3 - Prob. 26PCh. 13.3 - Prob. 27PCh. 13.3 - Prob. 28PCh. 13.3 - 13–29 A gas mixture at 350 K and 300 kPa has the...Ch. 13.3 - Prob. 30PCh. 13.3 - Prob. 31PCh. 13.3 - A rigid tank that contains 2 kg of N2 at 25C and...Ch. 13.3 - Prob. 33PCh. 13.3 - Prob. 34PCh. 13.3 - Prob. 35PCh. 13.3 - Prob. 36PCh. 13.3 - A 30 percent (by mass) ethane and 70 percent...Ch. 13.3 - Prob. 38PCh. 13.3 - Prob. 39PCh. 13.3 - Prob. 40PCh. 13.3 - Prob. 41PCh. 13.3 - Prob. 42PCh. 13.3 - Prob. 43PCh. 13.3 - Is the total internal energy of an ideal-gas...Ch. 13.3 - Prob. 45PCh. 13.3 - Prob. 46PCh. 13.3 - 13–47C Is the total internal energy change of an...Ch. 13.3 - Prob. 48PCh. 13.3 - Prob. 49PCh. 13.3 - The volumetric analysis of a mixture of gases is...Ch. 13.3 - Prob. 52PCh. 13.3 - Prob. 53PCh. 13.3 - Prob. 54PCh. 13.3 - Prob. 55PCh. 13.3 - Prob. 56PCh. 13.3 - An insulated tank that contains 1 kg of O2at 15C...Ch. 13.3 - Prob. 59PCh. 13.3 - Prob. 60PCh. 13.3 - Prob. 61PCh. 13.3 - Prob. 62PCh. 13.3 - Prob. 63PCh. 13.3 - Prob. 64PCh. 13.3 - Prob. 66PCh. 13.3 - Prob. 67PCh. 13.3 - Prob. 69PCh. 13.3 - A pistoncylinder device contains 6 kg of H2 and 21...Ch. 13.3 - Prob. 71PCh. 13.3 - Prob. 72PCh. 13.3 - Prob. 73PCh. 13.3 - Prob. 74PCh. 13.3 - Prob. 75PCh. 13.3 - Prob. 76PCh. 13.3 - Prob. 77PCh. 13.3 - Prob. 78PCh. 13.3 - Prob. 80PCh. 13.3 - Prob. 81PCh. 13.3 - Fresh water is obtained from seawater at a rate of...Ch. 13.3 - Prob. 83PCh. 13.3 - Prob. 84RPCh. 13.3 - The products of combustion of a hydrocarbon fuel...Ch. 13.3 - Prob. 89RPCh. 13.3 - Prob. 91RPCh. 13.3 - Prob. 92RPCh. 13.3 - A spring-loaded pistoncylinder device contains a...Ch. 13.3 - Prob. 94RPCh. 13.3 - Reconsider Prob. 1395. Calculate the total work...Ch. 13.3 - A rigid tank contains a mixture of 4 kg of He and...Ch. 13.3 - Prob. 97RPCh. 13.3 - Prob. 100RPCh. 13.3 - An ideal-gas mixture whose apparent molar mass is...Ch. 13.3 - 13–102 An ideal-gas mixture consists of 2 kmol of...Ch. 13.3 - An ideal-gas mixture consists of 2 kmol of N2and 4...Ch. 13.3 - Prob. 104FEPCh. 13.3 - Prob. 105FEPCh. 13.3 - An ideal-gas mixture consists of 3 kg of Ar and 6...Ch. 13.3 - Prob. 107FEPCh. 13.3 - Prob. 108FEPCh. 13.3 - Prob. 109FEPCh. 13.3 - Prob. 110FEP
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